GREPhysics.NET: GR9677 #81

GR9677 #81

Problem

GREPhysics.NET Official Solution

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Wave Phenomena $\Rightarrow$ }Beats

One remembers the relation for beats, i.e., $f_1-f_2=f_{beat}$ . Beat phenomenon occurs when two waves occur at nearly the same frequency.

Taking $f_0>\approx 73$ as the fundamental frequency, one deduces the harmonic to be $440/f_0\approx 6$ . $440-6f_0\approx 0.5$ , and thus the answer is choice (B).

One can derive the relation for beats by recalling the fact that one gets beat phenomenon when one superposes two sound waves of similar frequency $f_1\approx f_2$ , say, of the form $A \sin(2 \pi f_i t)$ ,

$f=A \sin(2 \pi f_1 t)+A \sin(2 \pi f_2 t)=2A\sin(2\pi(f_1+f_2)/2 t)\cos(2\pi(f_1-f_2)/2 t), <br />$

where to get beats phenomenon one must have $\cos(t (f_1-f_2)/2)=\pm 1 \Rightarrow 2\pi = (f_1-f_2)/2 t$ , and since there are two beats per period, one has $f_1-f_2=f_{beat}$ .

Alternate Solutions

ramparts 2009-10-01 12:43:38	Limits, limits, limits! No need to know about beat frequencies and all that fancy physics stuff to get this right (I didn't! :P). Just play the GRE game. One can figure pretty easily that if the string tensions are the same, the amplitude stays at 1, because then they're the same string and there's absolutely no reason for the wave to change. That eliminates D and E. Meanwhile, if the right side is EXTREMELY heavy, you expect the amplitude to be pretty darn small - so as $\mu_r$ goes to $\infty$ , the amplitude should go to zero. This eliminates A and C, leaving only B. There you go :) Reply to this comment
jmason86 2009-09-22 19:30:44	Basically the same solution but using some test taking strategy: ETS gave two answers with Harmonic = 6. The correct answer is PROBABLY one of these (since all the others are different). Eliminate (C) (D) and (E). 73*6 = 438. 440-438=2 but since the actual frequency is just over 73, the difference should be less than 2. Eliminate (A)... (B) remains. Reply to this comment

Comments

ramparts
2009-10-01 12:43:38

Limits, limits, limits! No need to know about beat frequencies and all that fancy *physics* stuff to get this right (I didn't! :P). Just play the GRE game. One can figure pretty easily that if the string tensions are the same, the amplitude stays at 1, because then they're the same string and there's absolutely no reason for the wave to change. That eliminates D and E. Meanwhile, if the right side is EXTREMELY heavy, you expect the amplitude to be pretty darn small - so as $\mu_r$ goes to $\infty$ , the amplitude should go to zero. This eliminates A and C, leaving only B. There you go :)

ramparts
2009-10-01 12:44:32

Ugh! Delete this. I was looking at problem 80. I was wondering why all the solutions talked about beat frequencies ;)

memorial
2010-07-08 08:19:18

wrong problem, buddy. we're on #81, not #80.

Reply to this comment

jmason86
2009-09-22 19:30:44

Basically the same solution but using some test taking strategy:

ETS gave two answers with Harmonic = 6. The correct answer is PROBABLY one of these (since all the others are different). Eliminate (C) (D) and (E).

73*6 = 438. 440-438=2 but since the actual frequency is just over 73, the difference should be less than 2. Eliminate (A)... (B) remains.

shak
2010-08-19 07:02:42

Can u please explain me, why are u dividing frequency of A4 to the frequency of D2 to find harmonics?
even , if then answer is close to 6 , but it is smaller than 6, and therefore answer should be 5,i.e lowest? thank you

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